Question #67465

Let V be the set of all functions that are twice differentiable in R and S={cosx,sinx,xcosx,xsinx}. a)Check that S is a linearly independent set over R.(Hint: Consider the equation a0cosx+a1sinx+a2xcosx+a3xsinx. Putx=0,π,π 2 ,π 4 ,etc.and solve for ai.) b) Let W=[S]and let T:V→V be the function defined by T(f(x))=d2 dx2(f(x))+2d dx(f(x)). Check that T is a linear transformation on V.
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Expert's answer

2017-04-18T13:53:05-0400

Answer on Question #67465 – Math – Linear Algebra

Question

Let VV be the set of all functions that are twice differentiable in R\mathbb{R} and


S={cosx,sinx,xcosx,xsinx}.S = \{\cos x, \sin x, x \cos x, x \sin x\}.


a) Check that SS is a linearly independent set over R\mathbb{R}. (Hint: Consider the equation


a0cosx+a1sinx+a2xcosx+a3xsinx.a_0 \cos x + a_1 \sin x + a_2 x \cos x + a_3 x \sin x.


Put x=0,π2,π4x = 0, \frac{\pi}{2}, \frac{\pi}{4} etc and solve for aia_i.)

b) Let V=[S]V = [S] and let T:VVT: V \to V be the function defined by T(f(x))=2x2f(x)+2xf(x)T\big(f(x)\big) = \frac{\partial^2}{\partial x^2} f(x) + 2\frac{\partial}{\partial x} f(x). Check that TT is a linear transformation on VV.

Solution

a) a0cosx+a1sinx+a2xcosx+a3xsinx=0a_0 \cos x + a_1 \sin x + a_2 x \cos x + a_3 x \sin x = 0.

For x=0x = 0:


a0cos0+a1sin0+a20cos0+a30sin0=0a0+0a1+0a2+0a3=0a0=0.\begin{array}{l} a_0 \cos 0 + a_1 \sin 0 + a_2 \cdot 0 \cdot \cos 0 + a_3 \cdot 0 \cdot \sin 0 = 0 \rightarrow \\ a_0 + 0 a_1 + 0 a_2 + 0 a_3 = 0 \rightarrow \\ a_0 = 0. \end{array}


For x=πx = \pi:


a1sinπ+a2πcosπ+a3πsinπ=00a1πa2+0a3=0a2=0.\begin{array}{l} a_1 \sin \pi + a_2 \cdot \pi \cdot \cos \pi + a_3 \cdot \pi \cdot \sin \pi = 0 \rightarrow \\ 0 a_1 - \pi a_2 + 0 a_3 = 0 \rightarrow \\ a_2 = 0. \end{array}


For x=π2x = \frac{\pi}{2}:


a1sinπ2+a3π2sinπ2=0a1+π2a3=0a1=π2a3.\begin{array}{l} a_1 \sin \frac{\pi}{2} + a_3 \cdot \frac{\pi}{2} \cdot \sin \frac{\pi}{2} = 0 \rightarrow \\ a_1 + \frac{\pi}{2} a_3 = 0 \rightarrow \\ a_1 = -\frac{\pi}{2} a_3. \end{array}


For x=π4x = \frac{\pi}{4}:


(π2a3)sinπ4+a3π4sinπ4=022(π2a3)+π422a3=0π2a3+π4a3=0π4a3=0a3=0,a1=π2a3=0.\begin{array}{l} \left(-\frac{\pi}{2} a_3\right) \sin \frac{\pi}{4} + a_3 \cdot \frac{\pi}{4} \cdot \sin \frac{\pi}{4} = 0 \rightarrow \\ \frac{2}{\sqrt{2}} \left(-\frac{\pi}{2} a_3\right) + \frac{\pi}{4} \cdot \frac{2}{\sqrt{2}} a_3 = 0 \rightarrow \\ -\frac{\pi}{2} a_3 + \frac{\pi}{4} a_3 = 0 \rightarrow \\ -\frac{\pi}{4} a_3 = 0 \rightarrow \\ a_3 = 0, \quad a_1 = -\frac{\pi}{2} a_3 = 0. \end{array}


Thus, over R\mathbb{R} the equation a0cosx+a1sinx+a2xcosx+a3xsinx=0a_0 \cos x + a_1 \sin x + a_2 x \cos x + a_3 x \sin x = 0 has only the trivial solution a0=a1=a2=a3=0a_0 = a_1 = a_2 = a_3 = 0, therefore SS is a linearly independent set over R\mathbb{R}.

b) Linear transformation is such that


T(kf)=kT(f),kRT(kf) = kT(f), \, k \in \mathbb{R}


and


T(f+g)=T(f)+T(g).T(f + g) = T(f) + T(g).


Since differentiation is a linear operation


(u+v)=u+v and (ku)=ku,(u + v)' = u' + v' \text{ and } (ku)' = ku',


transformation TT is also linear.

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